Doubts in SOlids !!

got it ! :P

one more ..

What is the maximum number of layers of atoms in closed packed planes that will lie within two imaginary parellel planes having a distance between them of 13 √2/3 R in the copper crystal (FCC) ?
COnsider the atoms to be within the parallel planes if their centres are on or within the two parellel planes.

is ans. five

Mere notes mein likha hai ki- Distance b/w A & B layers in a homogenous FCC packing(ABCABC..type)
= 2r√2/3.

So, tht way, ans shud be 6.....

Distance koi derive karo yaar, dimaag nahi chal rahaa...

I can prove it
yeah correct Avik, h = 2a √(2/3) = 4r√(2/3)

Layer A contians the corner atoms and the face centre atoms where as layer B contains within the unit cell
The distance b/w two succesive same layers is the height of the unit cell. All the layers are equivalent frm each other. All the atoms shuld be same. The corner atoms are touching each other along the edges of the regualr hexagon.

Since the regular hexagon can be divided into six equilateral triangles, it can be inferred that the corner atoms are making contact with the face centre atoms.

The atom of alyer B is located over the centroid of the equilateral triangle formed by three aotms of layer A. Atoms of layer B is making contact with all the three atoms of layer A.

AB = 2 r = AE
DE = R
In triangle DEA, DA= √(AE² - DE²)
= √[(2r)² - r ² ]
= √3 r

Now, ACDC = 2
So, AC = (2/3) DA = (2/3) x √3 r = 2r / √3

Now, In triangle ABC, BC = h/2 = √[(AB)²- (AC)²]
= √[(2r)² - (2r/√3) ² ]
Therefore, h/2 = 2r √2/3
h = 4r √2/3

Derivitaion for packing fraction :-

V cell = (Area of base) X height
= (6 X Area of equilateral triangle) X h
=6 (√3a² / 4) h
Putting h = 2a √2/3
= 3 √2 a ³
= 3√2 . 8 r ³
Therefore, V cell = 24 √2 r ³

V cell is dependent only on the radius of the atom,
Therefore, P.F = V atomV cell = 8 ∩ / (24 √2) = ∩ / 3√2

P.F = 0.74

Tushar why hcp ???????

Ans is neither 5 nor 6 .. it's ........"7"

And yea WHY HCP ???